2

Here is a problem I can not solve :

Let $X(x, y)$ be a $C^{\infty}$ vector field defined on $\mathbb{R}^2$ such that $X(x, y) = (y + a(x, y), x+ b(x, y))$ where $a, b : \mathbb{R}^2 \to \mathbb{R}$ are bounded and $a(0) = b(0) = 0$ and $da(0) = db(0) = 0$.

We assume there exist periodic points $ p_n = (x_n, y_n) \to (0, 0)$ (and $p_n \neq 0$) such that the solution of $ (x',y') =X(x,y)$ and $(x(0),y(0))=p_n$ is $T_n$ periodic. I want to show that $T_n \to \infty$

What I've done :

  1. We have $\|(x',y')\| \le \| (x,y)\| + C $ with $C$ a constant so $\phi ^t$ is defined on $\mathbb{R}$

  2. $0$ is an hyperbolic point

  3. By the Hartman Grobman theorem I proved that there is a $\varepsilon$ such that the ball $B(0,\varepsilon)$ does not contain any periodic orbit.

Argue by contradiction. Suppose there is an extraction $\phi$ such that $T_{\phi (n)}$ is bounded by $T_{\infty}$. Then I named $L_n$ the length of the orbit $\phi ^t (p_n)$ and I want to show that if $n$ is large enough then $L_n\le \varepsilon $ :

$$ L_n= \int_0^{T_n} \sqrt{x'(t)^2+y'(t)^2} dt \le \int_0^{T_\infty} \sqrt{x'(t)^2+y'(t)^2} dt $$

What can I do to finish ?

Thank you.

bartgol
  • 6,231
M.LTA
  • 1,294
  • Well, you can use Grobman-Hartman theorem further. The "passage time" near the saddle (if you define it accurately) is the same for the linearized system and for the original non-linear system. And it's almost a trivial fact that the closer you are to the saddle, the more time you need to pass near it. Since period of periodic orbit is greater than this passage time and passage time tends to infinity, period tends to infinity too. – Evgeny Oct 11 '15 at 21:37
  • Can you be more precise please ? I don't find your trivial fact so trivial I was not able to prove it. And I don't know why the passage time is the same for the linearized, Hartman Grobman is not saying this in my opinion. Thank you ! – M.LTA Oct 12 '15 at 08:08
  • 1
    Sorry, I forgot to add that it's trivial fact for linear saddle with diagonalized matrix :) I think it became much easier, isn't it? However, I understand that my comment was too sketchy and I'll write the full answer later. – Evgeny Oct 12 '15 at 08:25

1 Answers1

1

This picture illustrates my idea from comment. Map $h$ denotes the conjugating homeomorphism between linear and non-linear flows and $G$ is some closed domain.

Passage time illustration

As you've already proved, there's a neighbourhood of saddle which contains no periodic orbits. Also, any periodic orbit that is very close to saddle has non-empty intersection with $h(G)$. Combining these two facts,we get that periodic orbit necessarily leaves $h(G)$ and its period is greater or equal to the amount of time spent in $h(G)$. But, if trajectory $\gamma$ leaves $h(G)$ then its preimage $h^{-1}(\gamma)$ (where it is defined, of course) leaves $G$. From conjugacy also follows that amount of time that $h^{-1}(\gamma)$ spends in $G$ is the same.

It's easy to derive this time for nice choice of $G$. For example, you can choose $G = \lbrace (x, y) \colon \vert x \vert < \delta, \vert y \vert < \delta \rbrace$. Suppose that periodic orbit enters through point $(\delta, y_0)$, $y_0 > 0$. By $\tau (y_0)$ denote the time at which trajectory intersects line $y = \delta$. Since the equations for linearized system are uncoupled, the solution for $y$ that passes through $(\delta, y_0)$ is $y(t) = y_0 e^{\alpha t}$ (for some $\alpha > 0$). The time $\tau(y_0)$ equals $\frac{1}{\alpha} \ln \frac{\delta}{y_0}$. Clearly, $\tau(y_0)$ tends to $+\infty$ as $y_0$ tends to zero.

Evgeny
  • 5,755